MathExtremist Joined: Aug 31, 2010
• Posts: 6526
September 6th, 2010 at 9:28:44 PM permalink
Quote: Wizard

Okay, suppose I have a fair coin and I say "I wrote one positive number on each side. The larger number is twice the smaller number." I then flip the coin. What is the probability it lands on the side with the larger number?

50%, and it doesn't matter what's written on the coin. But "positive numbers" are infinite...
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
Wizard Joined: Oct 14, 2009
• Posts: 23448
September 6th, 2010 at 9:38:17 PM permalink
Quote: MathExtremist

50%, and it doesn't matter what's written on the coin. But "positive numbers" are infinite...

Since the "positive numbers" part bothers you, what if I change it to say "I wrote one positive number on each side. The larger number is twice the smaller number. The larger amount is less than the amount of money I own."
It's not whether you win or lose; it's whether or not you had a good bet.
MathExtremist Joined: Aug 31, 2010
• Posts: 6526
September 6th, 2010 at 9:45:37 PM permalink
Quote: Wizard

Since the "positive numbers" part bothers you, what if I change it to say "I wrote one positive number on each side. The larger number is twice the smaller number. The larger amount is less than the amount of money I own."

It's a fair coin, so it's still 50% to land on the larger value (on 2X, just for consistency).
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
Wizard Joined: Oct 14, 2009
• Posts: 23448
September 6th, 2010 at 10:16:15 PM permalink
Quote: MathExtremist

It's a fair coin, so it's still 50% to land on the larger value (on 2X, just for consistency).

Okay, good, we've come this far. Now suppose instead of writing the amounts on a coin I write them on two pieces of paper. Then I put each paper in an envelope. Then I glue the envelopes together. Then I throw of set of envelopes off the roof of a 12-story building. After landing, what are the odds that the face up envelope has the larger (2x) amount of money inside?
It's not whether you win or lose; it's whether or not you had a good bet.
MathExtremist Joined: Aug 31, 2010
• Posts: 6526
September 6th, 2010 at 10:27:36 PM permalink
Quote: Wizard

Okay, good, we've come this far. Now suppose instead of writing the amounts on a coin I write them on two pieces of paper. Then I put each paper in an envelope. Then I glue the envelopes together. Then I throw of set of envelopes off the roof of a 12-story building. After landing, what are the odds that the face up envelope has the larger (2x) amount of money inside?

I'm not disagreeing that there's a 50% chance of picking the envelope with 2X. I'm disagreeing that that means the *other* envelope has an equal chance (50/50) of being half or double the value of the first one. Going back to your list:
Quote: Wizard

5. The odds the other envelope has twice the money of mine is 50%

is not true for all possible values that can be in your envelope.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
Wizard Joined: Oct 14, 2009
• Posts: 23448
September 6th, 2010 at 10:34:51 PM permalink
Quote: MathExtremist

I'm not disagreeing that there's a 50% chance of picking the envelope with 2X. I'm disagreeing that that means the *other* envelope has an equal chance (50/50) of being half or double the value of the first one. Going back to your list: is not true for all possible values that can be in your envelope.

Good, this narrows down the point of departure some more.

So, if you agree there is a 50% chance you chose the larger (2x) envelope before opening it, then would you agree the odds are 50% that you chose the smaller envelope (x) as well?
It's not whether you win or lose; it's whether or not you had a good bet.
MathExtremist Joined: Aug 31, 2010
• Posts: 6526
September 6th, 2010 at 10:45:35 PM permalink
Quote: Wizard

Good, this narrows down the point of departure some more.

So, if you agree there is a 50% chance you chose the larger (2x) envelope before opening it, then would you agree the odds are 50% that you chose the smaller envelope (x) as well?

Of course. Maybe substitute "instead" for "as well", but I know what you meant.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
Wizard Joined: Oct 14, 2009
• Posts: 23448
September 6th, 2010 at 10:56:29 PM permalink
Quote: MathExtremist

Of course. Maybe substitute "instead" for "as well", but I know what you meant.

So, I think we can agree that:

1. The probability your envelope is the big one is 50%.
2. The probability your envelope is the small one is 50%.
3. The large envelope has twice the amount of the small envelope.

Now, can we say that if you chose the smaller envelope than the other one is the larger envelope?
It's not whether you win or lose; it's whether or not you had a good bet.
MathExtremist Joined: Aug 31, 2010
• Posts: 6526
September 6th, 2010 at 11:31:42 PM permalink
Quote: Wizard

So, I think we can agree that:

1. The probability your envelope is the big one is 50%.
2. The probability your envelope is the small one is 50%.
3. The large envelope has twice the amount of the small envelope.

Now, can we say that if you chose the smaller envelope than the other one is the larger envelope?

Yes.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
scotty81 Joined: Feb 4, 2010
• Posts: 185
September 7th, 2010 at 6:06:14 AM permalink
I'd like to propose we look at this a slightly different way in order to expose the flaw. It's difficult to see what is going on when you think in terms of only one coin, or one set of envelopes.

In order to simulate the paradox, let's say we have 100 coins. On each coin we write "X" on one side and either "2X" or ".5X" on the other side. To properly simulate the true probabilities, 50 of our coins have "2X" on one side, and 50 of our coins have ".5X" on one side. We then put the coins into a jar, reach in and choose one at random.

The coin represents a random set of two envelopes that are presented to us. We then flip the coin to choose the envelope. The envelope we choose is represented by what comes up on the coin.

There is a 50% chance that we will see an "X" as the result.

There is a 50% chance that we will NOT see an "X". In the event we do not see the "X", then there is a 50% chance we will see ".5X", and a 50% chance we will see "2X".

Therefore, the overall probability of what we will see on the coin is:

50% we will see "X"
25% we will see ".5X"
25% we will see "2X"

Now, what is the expected value of not accepting this result, and taking the value of the other side of the coin?

If you are looking at the "X", then it is true that the EV is 25% of X if you take the other side (in terms of your net gain.) But, this only applies IF THE "X" SIDE IS THE SIDE THAT CAME UP. This is the fundamental source of the flaw; you can't assume that every time you flip the coin that "X" will come up.

If the ".5X" side comes up (25% chance this will happen), then the EV of taking the other side is 50% of X (turning .5X into X).

If the "2X" side comes up (25% chance this will happen), then the EV of taking the other side is -100% of X (turning 2X into X).

So, overall, you have a 50% chance of obtaining an EV of 25%
You have a 25% chance of obtaining an EV of 50%
You have a 25% chance of obtaining an EV of -100%

Total net EV for all cases: (25% * .5) + (50% * .25) + (-100% * .25) = 0% EV

*******************************

Now, let's look at this in terms of the steps outlined by the Wizard:

4) The odds I chose the smaller envelope is 50%.
5) The odds the other envelope has twice the money of mine is 50%.
6) The odds the other envelope has half the money of mine is 50%.

Actually, these are accurate statements. But, if you change the wording to be in line with what follows in the sequence, they are not. The correct wording should be:

4) The odds I chose the smaller envelope is 50%.
5) The odds I CHOSE the envelope with twice the money is 50%
6) The odds I CHOSE the envelope with half the money is 50%

In this context (the correct context), you can't have the odds add up to 150%. The total odds of your choice can only be 100%. The correct assumptions have to be:

4) The odds I chose the smaller envelope is 50%.
5) The odds I CHOSE the envelope with twice the money is 25%
6) The odds I CHOSE the envelope with half the money is 25%

The odds of the other envelope being 50% with twice the money, and 50% with half the money only applies IF the smaller envelope has been chosen, which is only true 50% of the time.

The Flaw is in assumning that you chose the smaller envelope 100% of the time, or in the example of the coin, that when you flip the coin that "X" is always the result that comes up.
Prediction is very difficult, especially about the future. - Niels Bohr